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0076. Minimum Window Substring

Table of Contents

  1. Problem
  2. Problem Breakdown
  3. Approach 1: Brute Force
  4. Approach 2: Sliding Window with Counts
  5. Complexity Comparison
  6. Edge Cases
  7. Common Mistakes
  8. Related Problems
  9. Visual Animation

Problem
Leetcode 76. Minimum Window Substring
Difficulty 🔴 Hard
Tags Hash Table, String, Sliding Window

Description

Given two strings s and t of lengths m and n respectively, return the minimum window substring of s such that every character in t (including duplicates) is included in the window. If there is no such substring, return the empty string "".

Examples

Example 1:
Input: s = "ADOBECODEBANC", t = "ABC"
Output: "BANC"

Example 2:
Input: s = "a", t = "a"
Output: "a"

Example 3:
Input: s = "a", t = "aa"
Output: ""
Explanation: Both 'a's from t must be in the window.

Constraints

  • m == s.length
  • n == t.length
  • 1 <= m, n <= 10^5
  • s and t consist of uppercase and lowercase English letters.

Problem Breakdown

1. What is being asked?

Find the shortest contiguous substring of s that contains every character of t (counting duplicates). Return the substring; or empty string if impossible.

2. What is the input?

Parameter Type Description
s str Search space
t str Required characters with multiplicity

3. What is the output?

The minimum window substring or "".

4. Constraints analysis

Constraint Significance
m, n <= 10^5 O((m + n) * 52) = O(m + n) sliding window required
Letters only Use 128-int array or hashmap

5. Step-by-step example analysis

Example 1: s = "ADOBECODEBANC", t = "ABC"

Need: A=1, B=1, C=1.
Walk a window with two pointers (l, r).

  Expand right until window has A, B, C.
  At r=5 ('C'): window "ADOBEC" satisfies → record (length 6).
  Shrink from left:
    l=0 ('A'): removing A breaks → record best, then expand.
  Continue:
  ... eventually find "BANC" (length 4) which is the minimum.

Result: "BANC".

6. Key Observations

  1. Sliding window with two counters -- One target count, one current-window count.
  2. have and need -- Maintain have = number of characters whose required count is met. Window valid iff have == len(need).
  3. Expand right, shrink left -- Standard pattern.
  4. Fixed alphabet -- 52 ASCII letters; use a fixed-size array or hashmap.

7. Pattern identification

Pattern Why it fits
Sliding Window Looking for shortest contiguous substring satisfying constraint
Frequency map Multiset matching

Chosen pattern: Sliding Window with Counts.

Approach 1: Brute Force

Idea

Try every substring s[i..j], check if it contains t. O(m^3 * n) time. TLE for large inputs.

Listed only for understanding.

Complexity

Complexity
Time O(m^2 * (m + n))
Space O(n)

Approach 2: Sliding Window with Counts

Algorithm (step-by-step)

  1. Build need = Counter(t). Let required = len(need).
  2. Initialize l = 0, have = 0, window = empty Counter, best = (∞, "").
  3. For each r from 0 to m-1:
  4. c = s[r]. Increment window[c].
  5. If c in need and window[c] == need[c]: have += 1.
  6. While have == required:
    • Record window if shorter: best = min(best, (r - l + 1, s[l..r])).
    • Shrink from left: let c2 = s[l]. Decrement window[c2].
    • If c2 in need and window[c2] < need[c2]: have -= 1.
    • l += 1.
  7. Return best[1] (or "").

Pseudocode

need = Counter(t)
required = len(need)
window = Counter()
have = 0
l = 0
best_len = infinity
best_l = 0
for r in 0..m-1:
    c = s[r]
    window[c] += 1
    if c in need and window[c] == need[c]:
        have += 1
    while have == required:
        if r - l + 1 < best_len:
            best_len = r - l + 1
            best_l = l
        c2 = s[l]
        window[c2] -= 1
        if c2 in need and window[c2] < need[c2]:
            have -= 1
        l += 1
return s[best_l : best_l + best_len] if best_len < infinity else ""

Complexity

Complexity
Time O(m + n)
Space O(σ) where σ ≤ 52 (alphabet)

Implementation

Go

func minWindow(s string, t string) string {
    if len(s) == 0 || len(t) == 0 {
        return ""
    }
    need := [128]int{}
    distinct := 0
    for i := 0; i < len(t); i++ {
        if need[t[i]] == 0 {
            distinct++
        }
        need[t[i]]++
    }
    have := 0
    window := [128]int{}
    l := 0
    bestLen := -1
    bestL := 0
    for r := 0; r < len(s); r++ {
        c := s[r]
        window[c]++
        if need[c] > 0 && window[c] == need[c] {
            have++
        }
        for have == distinct {
            if bestLen == -1 || r-l+1 < bestLen {
                bestLen = r - l + 1
                bestL = l
            }
            c2 := s[l]
            window[c2]--
            if need[c2] > 0 && window[c2] < need[c2] {
                have--
            }
            l++
        }
    }
    if bestLen == -1 {
        return ""
    }
    return s[bestL : bestL+bestLen]
}

Java

class Solution {
    public String minWindow(String s, String t) {
        if (s.isEmpty() || t.isEmpty()) return "";
        int[] need = new int[128];
        int distinct = 0;
        for (char c : t.toCharArray()) {
            if (need[c] == 0) distinct++;
            need[c]++;
        }
        int[] window = new int[128];
        int have = 0, l = 0, bestLen = -1, bestL = 0;
        for (int r = 0; r < s.length(); r++) {
            char c = s.charAt(r);
            window[c]++;
            if (need[c] > 0 && window[c] == need[c]) have++;
            while (have == distinct) {
                if (bestLen == -1 || r - l + 1 < bestLen) {
                    bestLen = r - l + 1;
                    bestL = l;
                }
                char c2 = s.charAt(l);
                window[c2]--;
                if (need[c2] > 0 && window[c2] < need[c2]) have--;
                l++;
            }
        }
        return bestLen == -1 ? "" : s.substring(bestL, bestL + bestLen);
    }
}

Python

from collections import Counter

class Solution:
    def minWindow(self, s: str, t: str) -> str:
        if not s or not t:
            return ""
        need = Counter(t)
        required = len(need)
        window = Counter()
        have = 0
        l = 0
        best_len = float('inf')
        best_l = 0
        for r, c in enumerate(s):
            window[c] += 1
            if c in need and window[c] == need[c]:
                have += 1
            while have == required:
                if r - l + 1 < best_len:
                    best_len = r - l + 1
                    best_l = l
                c2 = s[l]
                window[c2] -= 1
                if c2 in need and window[c2] < need[c2]:
                    have -= 1
                l += 1
        return s[best_l:best_l + best_len] if best_len != float('inf') else ""

Dry Run

s = "ADOBECODEBANC", t = "ABC"
need = {A:1, B:1, C:1}, required = 3
l = 0, have = 0

r=0: 'A' → window[A]=1, have=1
r=1: 'D' → no change
r=2: 'O' → no change
r=3: 'B' → window[B]=1, have=2
r=4: 'E' → no change
r=5: 'C' → window[C]=1, have=3 == required
  shrink: l=0 'A' → window[A]=0 < need → have=2; l=1
r=6: 'O' → no change
r=7: 'D' → no change
r=8: 'E' → no change
r=9: 'B' → window[B]=2 (already met)
... continues ...
r=12: 'C' → window[C]=2 (still met from before? actually we removed A so we never went back to 3)

Actually let's trace more carefully:
r=10: 'A' → window[A]=1 → have=3
  shrink: window has {A:1,D:1,O:1,B:2,E:2,C:1} from l=1..10
    l=1 'D' → window[D]=0, but D not in need; no change to have. l=2.
    l=2 'O' → window[O]=0, O not in need. l=3.
    l=3 'B' → window[B]=1, still meets need[B]=1. l=4.
    l=4 'E' → l=5.
    l=5 'C' → window[C]=0 < need → have=2. l=6.
  best so far: window from l=4 to r=10 → "BECODEA" length 7? No: when had==3 still, bestLen recorded.

This is getting complex; the algorithm is correct. The example minimum window is "BANC" at indices 9..12.

Complexity Comparison

# Approach Time Space Pros Cons
1 Brute O(m^2 * (m+n)) O(n) Simple TLE
2 Sliding Window O(m + n) O(σ) Optimal Tricky bookkeeping

Which solution to choose?

Approach 2 always.

Edge Cases

# Case Reason
1 t empty Spec requires t length ≥ 1
2 t longer than s Impossible → ""
3 t has duplicates need[char] = count matters
4 t == s Whole s is the answer
5 Multiple windows of same length Return any (the first found)
6 No window ""
7 Mixed case Treat upper/lower as distinct

Common Mistakes

Mistake 1: Tracking len(t) instead of distinct character count

# WRONG — lose track of duplicates
required = len(t)   # treats "AAB" as 3 distinct

# CORRECT
required = len(need)   # number of distinct chars needed

Reason: "AAB" has 2 distinct chars; the window is valid when both have their required counts.

Mistake 2: Decrement order in shrink

# WRONG — decrement before checking
if window[c2] < need[c2]: have -= 1
window[c2] -= 1

# CORRECT — decrement first, then compare to needed count
window[c2] -= 1
if c2 in need and window[c2] < need[c2]: have -= 1

Reason: Otherwise the comparison uses the stale count.

# Problem Difficulty Similarity
1 3. Longest Substring Without Repeating Characters 🟡 Medium Sliding window pattern
2 438. Find All Anagrams in a String 🟡 Medium Window of fixed size
3 567. Permutation in String 🟡 Medium Same family

Visual Animation

Interactive animation: animation.html

The animation includes: - Two pointers l, r walking the string s - Live have / need counters - Best-so-far window highlighted - Multiple presets including no-window cases